BTW, if you need to change array later, use exchangeArray() method. Good to know when you are writing a class that extends ArrayObject()
AFAIK, exchangeArray() doesn't return anything.
<?php
$a = array('one', 'two', 'three');
$ao = new ArrayObject($a);
foreach ($ao as $element) {
echo $element . ' '; // one two three
}
$b = array('four', 'five', 'six');
$ao->exchangeArray($b); // returns null
foreach ($ao as $element) {
echo $element . ' '; // four five six
}
?>
ArrayObject::__construct
(PHP 5, PHP 7, PHP 8)
ArrayObject::__construct — Создаёт новый объект массива
Описание
public ArrayObject::__construct(array|object
$array = [], int $flags = 0, string $iteratorClass = ArrayIterator::class)Создаёт новый объект (object) массива.
Список параметров
-
array -
flags -
Флаги для управления поведением объекта ArrayObject. Смотрите ArrayObject::setFlags().
-
iteratorClass -
Указывает класс, который будет использоваться в качестве итератора объекта ArrayObject. Класс должен реализовать интерфейс ArrayIterator.
Примеры
Пример #1 Пример использования ArrayObject::__construct()
<?php
$array = array('1' => 'one',
'2' => 'two',
'3' => 'three');
$arrayobject = new ArrayObject($array);
var_dump($arrayobject);
?>Результат выполнения приведённого примера:
object(ArrayObject)#1 (3) {
[1]=>
string(3) "one"
[2]=>
string(3) "two"
[3]=>
string(5) "three"
}
add a note
User Contributed Notes 4 notes
german dot rumm at gmail dot com ¶
16 years ago
ashley at nospam dot zincdigital dot com ¶
15 years ago
The great confusion with this class is in its naming. ArrayObject infers it will behave as an Array and as an Object. It won't. It behaves as an array. It would better be called ArrayType. You can, with some work, get it to work both as an object and as an array, but that is up to you.
Grigori Kochanov ¶
18 years ago
As Marcus explained, the flag ArrayObject::SPL_ARRAY_AS_PROPS means the array element may be used as a property if there is no conflict with visible properties.
If there are visible properties in the class, the array element will not overwrite it's value.
<?php
class Rules extends ArrayObject {
public $len = 1;
function __construct($array){
parent::__construct($array,ArrayObject::ARRAY_AS_PROPS);
$this['len'] = 2;
}
}
$x = new Rules(array(1,2));
echo $x->len;
?>
Result: 1
<?php
class Rules extends ArrayObject {
private $len = 1;
function __construct($array){
parent::__construct($array,ArrayObject::ARRAY_AS_PROPS);
$this['len'] = 2;
}
}
$x = new Rules(array(1,2));
echo $x->len;
?>
Result: 2
agalkin at agalkin dot ru ¶
17 years ago
Note that the first argument to ArrayObject::__construct, the initial array, is passed by reference. Nevertheless, modification of the array doesn't modify the object, so it may cause unexpected behaviour.
<?php
$array = array('foo' => 'initial');
$obj = new ArrayObject($array);
// array was passed by reference:
$obj['foo'] = 'modified';
var_dump($array); // foo => modified
// but it doesn't work backwards:
$array['foo'] = 'modified_again';
var_dump($obj); // foo => modified
var_dump($array); // foo => modified_again
?>
